Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get:

$\frac{12+18y}{(1-3x{)}^{2}}$

But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:

$\frac{48}{(1-3x{)}^{3}}$

I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get:

$\frac{12+18y}{(1-3x{)}^{2}}$

But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:

$\frac{48}{(1-3x{)}^{3}}$

I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?